odeAdamsKrogh¶

Solves an initial-value problem for a system of ordinary differential equations of order one or two using a variable order Adams method.

Synopsis¶

odeAdamsKrogh (t, tend, ido, y, hidrvs, fcn)

Required Arguments¶

float t (Input/Output)

On input, t contains the initial independent variable value. On output, t is replaced by tend unless error conditions arise. See ido for details.

float tend (Input)

Value of t = tend where the solution is required.

int ido (Input/Output)

Flag indicating the state of the computation.

`ido`	State
1	Initial entry input value.
2	Normal re-entry input value. On output, if `ido` = 2 then the integration is finished. If the integrator is called with a new value for `tend`, the integration continues. If the integrator is called with `tend` unchanged, an error message is issued.
3	Input value to use on final call to release workspace.
>3	Output value that indicates that a fatal error has occurred.

The initial call is made with ido = 1. The function then sets ido = 2, and this value is used for all but the last call that is made with ido = 3. This final call is only used to release workspace which was automatically allocated by the initial call with ido = 1.

float y[] (Input/Output)

An array of length k containing the dependent variables, y(t), and first derivatives, if any. k will be the sum of the orders of the equations in the system of equations to solve, that is, the sum of the elements of eqOrder. On input, y contains the initial values, \(y(t_0)\) and \(y'(t_0)\) (if needed). On output, y contains the approximate solution, \(y(t)\). For example, for a system of first order equations, y[i‑1] is the i‑th dependent variable. For a system of second order equations, y[2i‑2] is the i‑th dependent variable and y[2i‑1] is the derivative of the i‑th dependent variable. For systems of equations in which one or more equations is of order 2, optional argument eqOrder must be used to denote the order of each equation so that the derivatives in y can be identified. By default it is assumed that all equations are of order 1 and y contains only dependent variables.

float hidrvs[] (Output)

An array of length neq containing the highest order derivatives at the point y.

void fcn (neq, ido, t, y[], hidrvs) (Input)

User-supplied function to evaluate derivatives.

Arguments

int neq (Input): Number of differential equations in the system of equations to solve.
int ido (Input): Flag indicating the state of the computation. This flag corresponds to the ido argument described above. If fcn has complicated subexpressions, which depend only weakly or not at all on y then these subexpressions need only be computed when ido = 1 and their values then reused when ido = 2.
float t (Input): Independent variable, t.
float y[] (Input): An array of length k containing the dependent variable values, y, and first derivatives, if any. k will be the sum of the orders of the equations in the system of equations to solve.
float hidrvs[] (Output): An array of length neq containing the values of the highest order derivatives evaluated at (t, y).

Optional Arguments¶

eqOrder, int (Input)

An array of length neq specifying the orders of the equations in the system of equations to solve. The elements of eqOrder can be 1 or 2. eqOrder must be used with argument y to define systems of mixed or higher order.

Default: eqOrder = [1,1,1,…,1].

eqErr, float (Input)

An array of length neq specifying the error tolerance for each equation. Let e(i) be the error tolerance for equation i for i = 0,…, neq ‑1. Then

Value	Explanation
\(e(i) > 0\)	Implies an absolute error tolerance of \(e(i)\) is to be used for equation i.
\(e(i) = 0\)	Implies no error checking is to be performed for equation i.
\(e(i) < 0\)	Implies a relative error test is to be performed for equation i. In this case, the base error tolerance used will be \(\|e(i)\|\) and the relative error factor used will be \((15/16 * \|e(i)\|)\). Thus the actual absolute error tolerance used will be \(\|e(i)\| \times (15/16 \times \|e(i)\|)\).

Default: An absolute error tolerance of 1.e‑5 is used for single precision and 1.e‑10 for double precision for all equations.

stepsizeInc, float (Input)

Factor used for increasing the stepsize. One should set stepsizeInc such that 9/8 <= stepsizeInc <= 4.

Default: stepsizeInc = 2.0.

stepsizeDec, float (Input)

Factor used for decreasing the stepsize. One should set stepsizeDec such that 1/4 <= stepsizeDec <= 7/8.

Default: stepsizeDec = 0.5.

minStepsize, float (Input)

Absolute value of the minimum stepsize permitted.

Default: minStepsize = 10.0/machine(2).

maxStepsize, float (Input)

Absolute value of the maximum stepsize permitted.

Default: maxStepsize = machine(2).

Description¶

odeAdamsKrogh is based on the JPL Library routine SIVA. odeAdamsKrogh uses a variable order Adams method to solve the initial value problem

\[\begin{split}\left. \begin{array}{l} \tfrac{dy_i}{dt} = f_i \left(t,y_1,y_2, \dots, y_{\mathit{neq}}\right) \\ y_i\left(t_0\right) = \eta_i \end{array} \right\}, i = 1,2, \ldots, \mathit{neq}\end{split}\]

or more generally

\[z_i^{\left(d_i\right)} = f_i(t,y),y\left(t_0\right) = \eta_0, i = 1,2, \ldots, \mathit{neq},\]

where y is the vector

\[\left(z_1, z'_1, \ldots, z_1^{\left(d_1-1\right)},z_2, \ldots, z_{\mathit{neq}}^{\left(d_{\mathit{neq}}-1\right)}\right),\]

\(z_i^{(k)}\) is the kth derivative of \(z_i\) with respect to t, \(d_i\) is the order of the ith differential equation, and η is a vector with the same dimension as y.

Note that the systems of equations solved by odeAdamsKrogh can be of order one, order two, or mixed order one and two.

See “Changing Stepsize in the Integration of Differential Equations Using Modified Divided Differences,” Krogh (1974).

Examples¶

Example 1¶

In this example a system of two equations of order two is solved.

\[Y''_1 = -Y_1 / \left(\left(Y_1^2+Y_2^2\right)^{\frac{3}{2}}\right)\]

\[Y''_2 = -Y_2 / \left(\left(Y_1^2+Y_2^2\right)^{\frac{3}{2}}\right)\]

The initial conditions are

\[Y_1(0) = 1.0, Y'_1(0) = 0.0, Y_2(0) = 0.0, Y'_2(0) = 1.0\]

Since the system is of order two, optional argument eqOrder must be used to specify the orders of the equations. Also, because the system is of order two, y[0] contains the first dependent variable, y[1] contains the derivative of the first dependent variable, y[2] contains the second dependent variable, and y[3] contains the derivative of the second dependent variable.

from __future__ import print_function
from numpy import *
from pyimsl.math.odeAdamsKrogh import odeAdamsKrogh
from pyimsl.math.constant import constant


def fcn(neq, ido, t, y, hidrvs):
    tp = y[0] * y[0] + y[2] * y[2]
    tp = 1.0e0 / (tp * sqrt(tp))
    hidrvs[0] = -y[0] * tp
    hidrvs[1] = -y[2] * tp


neq = 2
ido = [1]
t = [0.0]
y = [1.0, 0, 0, 1.0]
hidrvs = []
korder = [2, 2]

# Write Title
print("   T         Y1/Y2          Y1P/Y2P         Y1PP/Y2PP")

# Integrate ODE
iend = 0
delta = 2.0 * constant("PI")
for k in range(5):
    iend += 1
    tend = t[0] + delta
    if (tend > 20.0):
        tend = 20.0
    odeAdamsKrogh(neq, t, tend, ido, y, hidrvs, fcn,
                  eqOrder=korder)
    if (iend < 5):
        print("%8.4f %10.4f %15.4f %15.4f" %
              (t[0], y[0], y[1], hidrvs[0]))
        print("        %12.4f %15.4f %15.4f" %
              (y[2], y[3], hidrvs[1]))
    if (iend == 4):
        ido[0] = 3

Output¶

   T         Y1/Y2          Y1P/Y2P         Y1PP/Y2PP
  6.2832     1.0000         -0.0000         -1.0000
              0.0000          1.0000         -0.0000
 12.5664     1.0000         -0.0000         -1.0000
              0.0000          1.0000         -0.0000
 18.8496     1.0000         -0.0000         -1.0000
              0.0000          1.0000         -0.0000
 20.0000     0.4081         -0.9129         -0.4081
              0.9129          0.4081         -0.9129

Example 2¶

This contrived example illustrates how to use odeAdamsKrogh to solve a system of equations of mixed order.

The height, y(t), of an object of mass m above the surface of the Earth can be modeled using Newton’s second law as:

\[my'' = -mg - ky'\]

or

\[y'' = -g-(k/m)y'\]

where ‑mg is the downward force of gravity and ‑ky’ is the force due to air resistance, in a direction opposing the velocity. If the object is a meteor, the mass, m, and air resistance, k, will decrease as the meteor burns up in the atmosphere. The mass is proportional to \(r^3\) (r = radius) and the air resistance, presumably dependent on the surface area, may be assumed to be proportional to \(r^2\), so that \(k/m = k_0/r\). The rate at which the meteor’s radius decreases as it burns up may depend on r, on the velocity y’, and, since the density of the atmosphere depends on y, on y itself. However, we will construct a very simple model where the rate is just proportional to the square of the velocity,

\[r' = -c_0\left(y'\right)^2\]

We solve (1) and (2), with \(k_0 = 0.005\), \(c_0 = 10^{-8}\), \(g = 9.8\) and initial conditions \(y(0) = 100,000 \text{ meters}\), \(y'(0) = ‑1000 \text{ meters/second}\), \(r(0) = 1 \text{ meter}\).

from __future__ import print_function
from numpy import *
from pyimsl.math.odeAdamsKrogh import odeAdamsKrogh


def fcn(neq, ido, t, y, hidrvs):
    hidrvs[0] = -9.8 - .005 / y[2] * y[1]
    hidrvs[1] = -1.e-8 * y[1] * y[1]


neq = 2
ido = [1]
t = [0.0]
y = [100000.0, -1000.0, 1]
korder = [2, 1]
eqnerr = [.003, .003]
hidrvs = []

# Write Title
print("   T         Y1/Y2          Y1P/Y2P         Y1PP/Y2PP")

# Integrate ODE
iend = 0
delta = 10.0
for k in range(6):
    iend += 1
    tend = t[0] + delta
    if (tend > 50.0):
        tend = 50.0
    odeAdamsKrogh(neq, t, tend, ido, y, hidrvs, fcn,
                  eqOrder=korder,
                  eqErr=eqnerr)
    if (iend < 6):
        print("%8.4f  %10.4f  %12.4f %15.4f" %
              (t[0], y[0], y[1], hidrvs[0]))
        print("   %15.4f                   %15.4f" %
              (y[2], hidrvs[1]))
    if (iend == 5):
        ido[0] = 3

Output¶

   T         Y1/Y2          Y1P/Y2P         Y1PP/Y2PP
 10.0000  89773.0459    -1044.0096         -3.9701
            0.8954                           -0.0109
 20.0000  79150.9980    -1078.6333         -2.9083
            0.7826                           -0.0116
 30.0000  68240.9574    -1101.0377         -1.5031
            0.6635                           -0.0121
 40.0000  57184.9223    -1106.9633          0.4253
            0.5413                           -0.0123
 50.0000  46178.1509    -1089.8294          3.1700
            0.4201                           -0.0119

Warning Errors¶

`IMSL_` `TOLERANCE_TOO_SMALL`	The requested error tolerance, # is too small. Using # instead.
`IMSL_` `RESTART`	The stepsize has been reduced too rapidly The integrator is going to do a restart.

Fatal Errors¶

`IMSL_ADJUST_STEPSIZE1`	The current step length = #, is less than the minimum steplength, “`minStepsize`” = #, at the conclusion of the starting phase of the integration. Decreasing “`minStepsize`” to a value less than or equal to # may help.
`IMSL_ADJUST_STEPSIZE2`	The integrator needs to take a step smaller than # in order to maintain the requested local error. Decreasing “`minStepsize`” to a value less than or equal to # may help.
`IMSL_INCORRECT_TEND`	Either the new output point precedes the last one or it has the same value. “`tend`” = #.
`IMSL_ADJUST_ERROR_TOLERANCE`	The step length, H = #, is so small that when Tn + H is formed, the result will be the same as `Tn`, where `Tn` is the base value of the independent variable. If this problem is not due to a nonintegrable singularity, it can probably be corrected by translating “`t`” so that it is closer to 0. Reducing the error tolerance for the equations through argument “`eqErr`” may also help with this problem.
`IMSL_ERROR_TOLERANCE`	A local error tolerance of zero has been requested.
`IMSL_ERROR_PREVIOUS`	A fatal error has occurred because of the error reported in the previous error message.
`IMSL_STOP_USER_FCN`	Request from user supplied function to stop algorithm. User flag = “#”.